Question

It is claimed that 15% of the ducks in a particular region have patent schistosome infection. Suppose that seven ducks are selected at random. Let X equal the number of ducks that are infected. (a) Assuming independence, how is X distributed? (b) Find (i) P(X ≥ 2), (ii) P(X = 1), and (iii) P(X ≤ 3).

Answer 1

Answer:

(a) X follows a Binomial distribution

(b) (i) P(X ≥ 2) = 0.28348

P(X = 1) = 0.39601

P(X ≤ 3) = 0.98800

Step-by-step explanation:

(a) In this situation, the variable X equal to the number of ducks that are infected follows a Binomial distribution because we have:

• n identical and independent events: The 7 ducks that are selected at random
• 2 possible results for every event: success and fail. We can call success if the duck is infected and fail if the duck is not infected.
• A probability p of success and 1-p of fail: There is a probability p equal to 15% that the ducks have the infection and a probability of (100%-15%) that they don't.

(b) So, the probability that X ducks are infected is calculated as:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

$P(x)=\frac{7!}{x!(7-x)!}*0.15^{x}*(1-0.15)^{7-x}$

Then, Probability P(X = 1) is equal to:

$P(1)=\frac{7!}{1!(7-1)!}*0.15^{1}*(1-0.15)^{7-1}=0.3960$

At the same way, probability P(X ≥ 2) is equal to:

P(X ≥ 2) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7)

P(X ≥ 2) = 0.2097 + 0.0617 + 0.0109 + 0.0011 + 0.00006 + 0.00002

P(X ≥ 2) = 0.28348

And probability P(X ≤ 3) is equal to:

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(X ≤ 3) = 0.3206 + 0.3960 + 0.2097 + 0.0617

P(X ≤ 3) = 0.988

Answer 2

Answers:

b)

i) 0.2834

ii) 0.3960

iii) 0.9880

Solution:

To solve this we need to use the binomial probability

P(X=k)=$(n k)* p^{k}*(1-p) ^{n-k}$

a)

X= number of ducks infected

n=7

p=15%=0.15

P(X)=$(7 x)* 0.15^{x}*(1-.15) ^{7-x}$ ; x=0,1,2,...,7

b)

First we need to calculate by the definition of binomial probability at k=0,1,2,3

P(X=0)=$(7 0)* 0.15^{0}*(1-0.15) ^{7-0}$ = 0.3206 ;

P(X=1)=$(7 1)* 0.15^{1}*(1-.15) ^{7-1}$ = 0.3960 ;

P(X=2)=$(7 2)* 0.15^{2}*(1-.15) ^{7-2}$ = 0.2097  ;

P(X=3)=$(7 3)* 0.15^{3}*(1-.15) ^{7-3}$ = 0.0617  ;

(i) Find P(X≥2)

Using the complement rule, we have that : P(A´ )= 1-P(A)

P(X≥2)= 1- P(X<2)

= 1- ((P(X=0)+P(X=1))

= 1- 0.3206-0.3960

=0.2834

(ii) Find P(X=1)

We have to evaluate the definition of binomial probaility at k=1

Then we have that

P(X=1)=$(7 1)* 0.15^{1}*(1-.15) ^{7-1}$ = 0.3960

(iii) Find P(X ≤ 3)

We have to use the addition tule for mutually exclusive events

Addition rule: P(A∪B)=P(A)+P(B)

P(≤3)= P(X=0)+ P(X=1) + P(X=2) + P(X=3)

=0.3206 + 0.3960+ 0.2097 + 0.0617

= 0.9880